11
$\begingroup$

I've been working on a problem where I ended up with the following nasty first ODE: $$\frac{(-1+e^{l^2})(b^2+l^2)}{l^2}\phi'(l)+\frac{b^2 \left(1-e^{l^2}+l^2\right)+l^2 \left(-2+2 e^{l^2}+l^2\right)}{l^3}\phi(l)=k\frac{1-2b^2-2l^2}{b^2 + l^2},$$ where $k \in \mathbb{R}$ and $b \in \mathbb{R_{\geq 0}}$.

One can somehow simplify the equation and write it as: $$\phi'+\frac{b^2 \left(1-e^{l^2}+l^2\right)+l^2 \left(-2+2 e^{l^2}+l^2\right)}{\left(-1+e^{l^2}\right)\left(b^2+l^2\right)l}\phi= k \frac{l^2 \left(1-2 b^2-2 l^2\right)}{\left(-1+e^{l^2}\right) \left(b^2+l^2\right)^2}.$$

Writting the above as $\phi'+h(l)\phi=g(l)$ the solution is given by: $$\phi(l) = e^{-\int_1^l h(\xi) d\xi} \int_1^l g(\zeta)~e^{\int_1^\zeta h(\xi) d\xi} d\zeta+C_1 e^{ -\int_1^l h(\xi) d\xi},$$

where $C_1$ is an integration constant.

The problem is in solving that big integral in the previous expression. I can't solve it and Mathematica tells me that it can't be put in terms of standard math functions. I was then hoping that somebody could help me find an approximate solution maybe using the method of matched asymptotic expansions since I know that $ \phi(l)=O(1/l^2)$ as $l \to \infty$. I also know from symmetry that $\phi(0)=0$.

I'm still not very good with approximation methods, nor know many, so I would be really grateful for any help you could give me.

$\endgroup$
10
  • 1
    $\begingroup$ Have you tried the math stackexchange? This site is for conceptual physics questions. Yours seems to be pure math. If you can physically motivate your problem that might lead to a useful approximation method. For instance, how do you know the leading behaviour of $\phi(l)$? What does it mean? $\endgroup$ Jun 12, 2013 at 14:40
  • $\begingroup$ I tried to put the problem within a physics background by editing the question but looking at the closing votes it wasn't enough. The behaviour of $\phi$, the gravitational field, for large $l$ - $l$ is some kind of distance - is because of the classical limit. The spacetime is asymptotically flat by construction so for large $l$: $\phi=O(1/l^2)$. $\endgroup$
    – PML
    Jun 12, 2013 at 14:50
  • $\begingroup$ Why not just make the approximation that $\pm1+\exp(l^{2})\approx\exp(l^{2})$ and $\pm l^{2}+\exp(l^{2})\approx\exp(l^{2})$? $\endgroup$ Jun 12, 2013 at 21:07
  • $\begingroup$ @AlexNelson Hum, in the sense of the asymptotic expansion? I mean, take those approximations for $l \ll 1$ and get a result. Then for $l\gg 1$ those are still very good approximations plus $b^2+l^2\approx l^2$. Then get the result and then do the matching? If so it's actually simple. I thought I couldn't do that because of how it's explained in the wiki article... $\endgroup$
    – PML
    Jun 12, 2013 at 23:29
  • $\begingroup$ @PML posting a bounty could be a good idea $\endgroup$
    – Dilaton
    Jun 13, 2013 at 23:50

1 Answer 1

3
$\begingroup$

$\dfrac{(-1+e^{l^2})(b^2+l^2)}{l^2}\phi'(l)+\dfrac{b^2(1-e^{l^2}+l^2)+l^2(-2+2e^{l^2}+l^2)}{l^3}\phi(l)=k\dfrac{1-2b^2-2l^2}{b^2+l^2}$

$\dfrac{(e^{l^2}-1)(l^2+b^2)}{l^2}\phi'(l)+\dfrac{(e^{l^2}-1)(2l^2-b^2)+l^2(l^2+b^2)}{l^3}\phi(l)=\dfrac{k}{l^2+b^2}-2k$

$\phi'(l)+\biggl(\dfrac{2l^2-b^2}{l(l^2+b^2)}+\dfrac{l}{e^{l^2}-1}\biggr)\phi(l)=\dfrac{kl^2}{(l^2+b^2)^2(e^{l^2}-1)}-\dfrac{2kl^2}{(l^2+b^2)(e^{l^2}-1)}$

I.F. $=e^{\int\bigl(\frac{2l^2-b^2}{l(l^2+b^2)}+\frac{l}{e^{l^2}-1}\bigr)dl}=e^{\frac{3}{2}\ln(l^2+b^2)-\ln l+\frac{1}{2}\ln(1-e^{-l^2})}=\dfrac{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}{l}$

$\therefore\biggl(\dfrac{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}{l}\phi(l)\biggr)'=\dfrac{kle^{-l^2}}{\sqrt{l^2+b^2}\sqrt{1-e^{-l^2}}}-\dfrac{2kl\sqrt{l^2+b^2}e^{-l^2}}{\sqrt{1-e^{-l^2}}}$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\dfrac{kle^{-l^2}}{\sqrt{l^2+b^2}\sqrt{1-e^{-l^2}}}dl-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\dfrac{2kl\sqrt{l^2+b^2}e^{-l^2}}{\sqrt{1-e^{-l^2}}}dl$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\dfrac{kle^{-l^2}}{\sqrt{l^2+b^2}}\sum\limits_{n=0}^\infty\dfrac{(2n)!e^{-nl^2}}{4^n(n!)^2}dl-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int2kl\sqrt{l^2+b^2}e^{-l^2}\sum\limits_{n=0}^\infty\dfrac{(2n)!e^{-nl^2}}{4^n(n!)^2}dl$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{kl(2n)!e^{-(n+1)l^2}}{4^n(n!)^2\sqrt{l^2+b^2}}dl-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{kl(2n)!\sqrt{l^2+b^2}e^{-(n+1)l^2}}{2^{2n-1}(n!)^2}dl$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!e^{-(n+1)u}}{2^{2n+1}(n!)^2\sqrt{u+b^2}}du-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!\sqrt{u+b^2}e^{-(n+1)u}}{4^n(n!)^2}du~(\text{Let}~u=l^2)$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!e^{-(n+1)(v^2-b^2)}}{4^n(n!)^2}dv-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!v^2e^{-(n+1)(v^2-b^2)}}{2^{2n-1}(n!)^2}dv~(\text{Let}~v=\sqrt{u+b^2})$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!e^{-(n+1)(v^2-b^2)}}{4^n(n!)^2}dv+\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!v}{4^n(n!)^2(n+1)}d(e^{-(n+1)(v^2-b^2)})$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!e^{-(n+1)(v^2-b^2)}}{4^n(n!)^2}dv+\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=0}^\infty\dfrac{k(2n)!ve^{-(n+1)(v^2-b^2)}}{4^n(n!)^2(n+1)}-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int e^{-(n+1)(v^2-b^2)}~d\left(\sum\limits_{n=0}^\infty\dfrac{k(2n)!v}{4^n(n!)^2(n+1)}\right)$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!e^{-(n+1)(v^2-b^2)}}{4^n(n!)^2}dv+\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=0}^\infty\dfrac{k(2n)!ve^{-(n+1)(v^2-b^2)}}{4^n(n!)^2(n+1)}-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!e^{-(n+1)(v^2-b^2)}}{4^n(n!)^2(n+1)}dv$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=0}^\infty\dfrac{k(2n)!ve^{-(n+1)(v^2-b^2)}}{4^nn!(n+1)!}+\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{kn(2n)!e^{(n+1)b^2}e^{-(n+1)v^2}}{4^n(n!)^2(n+1)}dv$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=0}^\infty\dfrac{k(2n)!ve^{-(n+1)(v^2-b^2)}}{4^nn!(n+1)!}+\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=1}^\infty\dfrac{\sqrt\pi kn(2n)!e^{(n+1)b^2}\text{erf}(\sqrt{n+1}v)}{2^{2n+1}(n!)^2(n+1)^{\frac{3}{2}}}+\dfrac{Cl}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=0}^\infty\dfrac{k(2n)!\sqrt{l^2+b^2}e^{-(n+1)l^2}}{4^nn!(n+1)!}+\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=1}^\infty\dfrac{\sqrt\pi kn(2n)!e^{(n+1)b^2}\text{erf}(\sqrt{n+1}\sqrt{l^2+b^2})}{2^{2n+1}(n!)^2(n+1)^{\frac{3}{2}}}+\dfrac{Cl}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}$

$\endgroup$
4
  • 12
    $\begingroup$ Perhaps you should use some words... $\endgroup$ Jul 11, 2013 at 14:03
  • $\begingroup$ I think you may need a constant of integration. $\endgroup$ Jul 11, 2013 at 17:14
  • $\begingroup$ Sorry, I forgot to notice that $0<e^{-l^2}\leq1$ for $l\in\mathbb{R}$ , so in fact it should be relieved to expand $\dfrac{1}{\sqrt{1-e^{-l^2}}}$ . $\endgroup$ Aug 27, 2013 at 7:43
  • $\begingroup$ @AlexNelson What are words? $\endgroup$
    – TheDoctor
    Sep 22, 2014 at 0:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.